Question

What is the energy of the orbiting electron in a hydrogen atom with a radial quantum number of \(45 ?\)

Short answer

Answer: The energy of an electron in a hydrogen atom with a radial quantum number of 45 is approximately -6.42 x 10^-3 eV.

Step-by-step solution

Understand the energy equation of a hydrogen atom

The energy of an electron in a hydrogen atom depends on the principal quantum number (n) and it follows a formula: $$ E_n = -\frac{13.6 \ \text{eV}}{n^2}, \quad n \in \{1,2,3, ...\} $$ This equation helps us find the energy level at which the electron resides, given its principal quantum number. The negative sign represents that the energy of electron in an atom is less than zero, i.e., the electron is bound to the nucleus. Here, we are given the radial quantum number (45). To solve the problem, we must find the corresponding principal quantum number (n) first and then find the energy using the above formula.

Determine the principal quantum number (n) from the radial quantum number

In a hydrogen atom, we can determine the principal quantum number (n) by simply adding 1 to the radial quantum number: $$ n = l + 1 $$ Given the radial quantum number \(l = 45\), we can find the principal quantum number using the expression above: $$ n = 45 + 1 = 46 $$

Calculate the energy of the orbiting electron

Now that we have the principal quantum number (n = 46), we can use the energy equation for the hydrogen atom to find the energy of the orbiting electron. Plugging \(n = 46\) into the equation, we get: $$ E_{46} = -\frac{13.6 \ \text{eV}}{46^2} = -\frac{13.6 \ \text{eV}}{2116} \approx -6.42 \times 10^{-3} \ \text{eV} $$ The energy of the orbiting electron in a hydrogen atom with a radial quantum number of 45 is approximately \(-6.42 \times 10^{-3}\) eV.